Exercise 5.3.16

Proof. Let $F$ have characteristic $p$ and consider $f=x^p-x+a\in F[x]$.

(a)

$f^{\prime }=-1$, thus $f\wedge f^{\prime }=1$, so $f$ is separable.

(b)

Let $\alpha$ be a root of $f$ in some extension $L$ of $F$. Then $f(\alpha )={\alpha }^p-\alpha +a=0$, thus $$\begin{array}{llll}\hfill f(\alpha +1)& ={(\alpha +1)}^p-(\alpha +1)+a& \hfill & \\ \hfill & ={\alpha }^p+1-\alpha -1+a& \hfill & \\ \hfill & =0,& \hfill & \end{array}$$

$\alpha +1\in L$ is also a root of $f$.

(c)

So $\alpha ,\alpha +1,\dots ,\alpha +p-1$ are roots of $f$. These roots are distinct since $0,1,\dots ,p-1$ are the $p$ distinct elements of the prime subfield of $F$, isomorphic to ${𝔽}_p$, and identified with ${𝔽}_p$.

Thus $f$ is divisible by $(x-\alpha )\cdots (x-\alpha -p+1)$, of degree $p=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$. As both polynomials are monic,

$$\begin{array}{lll}\hfill f=(x-\alpha )(x-\alpha -1)\cdots (x-\alpha -p+1).& & \hfill \text{(1)}\end{array}$$

(d)

$\text{∙}$ $F(\alpha )$ contains $F$ and thus contains also ${𝔽}_p$. So $F(\alpha )$ contains $\alpha ,\alpha +1,\dots ,\alpha +p-1$, thus $F(\alpha )=F(\alpha ,\alpha +1,\dots ,\alpha +p-1)$.

$F(\alpha )$ is so the splitting field of $f$ by $\mathrm{\text{(}}\text{1<!-- tex4ht:ref: eq5.3.16:1 -->}\text{)}$. $F\subset F(\alpha )$ is a normal extension.

$\text{∙}$ The minimal polynomial $g$ of $\alpha$ over $F$ divides $f$, which has only simple roots, thus $g$ has only simple roots. So $\alpha$ is separable over $F$. By Theorem 5.3.15(a), $F\subset F(\alpha )$ is a separable extension.